Problem 1 - Precalculus: Mathematical for Calculus

Question

Write the form of the partial fraction decomposition of the function (as in Example 4). Do not determine the numerical values of the coefficients. $$\frac{1}{(x-1)(x+2)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{A}{x - 1} + \frac{B}{x + 2} \).

Step 1: Identify the Denominator

First, observe the denominator of the given function: \((x - 1)(x + 2)\). The denominator is already factored into linear terms \((x - 1)\) and \((x + 2)\).

Step 2: Set Up Partial Fraction Decomposition

For each unique factor in the denominator, we will set up a partial fraction. Because both \((x - 1)\) and \((x + 2)\) are linear factors, their numerators will be constants. Let the partial fraction decomposition be:\[ \frac{A}{x - 1} + \frac{B}{x + 2} \] where \(A\) and \(B\) are constants to be determined later.

Key Concepts

Linear Factors in Partial Fraction Decomposition

In mathematics, a linear factor is an expression of the form \(x - a\), where \(a\) is a constant. When dealing with partial fraction decomposition, recognizing linear factors is crucial.

This is because each linear factor in the denominator of a rational function contributes a term to the partial fraction decomposition that has a constant numerator.

• If your denominator has linear factors like \((x-1)\) and \((x+2)\), then each will become a separate fraction in the decomposition.
• The numerators of these fractions will be constants usually represented by letters like \(A\) and \(B\).

Partial fraction decomposition aims to break down complex rational expressions into simpler fractions, making them easier to integrate or simplify further.

Understanding Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. The rational function in this problem is \(\frac{1}{(x-1)(x+2)}\).

Here, the numerator is 1 which is a simple polynomial, and the denominator is the product of \(x-1\) and \(x+2\). Rational functions can often be tricky because their complexity comes from the interaction between the numerator and the denominator.

• They are common in calculus and algebra when working with limits, derivatives, and integrals.
• Understanding rational functions helps in recognizing how their behavior changes depending on the values of \(x\).

Thus, simplifying rational functions into partial fractions is a powerful technique for solving integrals or understanding the function's behavior more clearly.

Factored Denominators in Rational Functions

When factoring the denominator of a rational function, it becomes easier to handle and analyze.

The problem provided has a denominator already factored into linear terms: \((x-1)(x+2)\). This indicates that we have distinct linear factors, each contributing to the solution's partial fractions.

• Factoring the denominator simplifies the decomposition process.
• Factored denominators provide insight into the possible values of \(x\) that may make the function undefined (for example, \(x-1=0\) and \(x+2=0\)).

Once the denominator is factored as seen, it allows you to easily assign each factor a place in the partial fractions, each with its own constant numerator.